Please help! :) I need the answer quick

Mathematics

2 answers:

scoray [572]2 years ago

7 0

Answer:no clue

Step-by-step explanation:

i need the points im sorry

attashe74 [19]2 years ago

5 0

Answer:

x ≈ 2.3

Step-by-step explanation:

Given

x³ = 12 ( take the cube root of both sides )

x = $\sqrt[3]{12}$ ≈ 2.3 ( to 1 dec. place )

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George bought 2.5 pounds of each type of fruit shown on the sign apples $1.30/lb,grapes $1.65/lb,bananas $0.49/lb.what was the t

Stella [2.4K]

Answer:

$8.60

Step-by-step explanation:

Let’s find the cost for each item

Apples: 2.5 x 1.30 = 3.25

Grapes: 2.5 x 1.65 = 4.125

Bananas: 2.5 x 0.49 = 1.225

Now we will add all the costs which will give use our total!

3.25 + 4.125 + 1.225 = 8.6

So the answer will be $8.60

5 0

2 years ago

Prove that H c G is a normal subgroup if and only if every left coset is a right coset, i.e., aH = Ha for all a e G

Kaylis [27]

$\Rightarrow$

Suppose first that $H\subset G$ is a normal subgroup. Then by definition we must have for all $a\in H$ , $xax^{-1} \in H$ for every $x\in G$ . Let $a\in G$ and choose $(ab)\in aH$ ( $b\in H$ ). By hypothesis we have $aba^{-1} =abbb^{-1}a^{-1}=(ab)b(ab)^{-1} \in H$ , i.e. $aba^{-1}=c$ for some $c\in H$ , thus $ab=ca \in Ha$ . So we have $aH\subset Ha$ . You can prove $Ha\subset aH$ in the same way.

$\Leftarrow$

Suppose $aH=Ha$ for all $a\in G$ . Let $h\in H$ , we have to prove $aha^{-1} \in H$ for every $a\in G$ . So, let $a\in G$ . We have that $ha^{-1} =a^{-1}h'$ for some $h'\in H$ (by the hypothesis). hence we have $aha^{-1}=h' \in H$ . Because $a$ was chosen arbitrarily we have the desired .